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Induction Proofs

  1. Jun 10, 2015 #1
    1. The problem statement, all variables and given/known data
    Prove that ∑nj=0 n C r = 2n


    2. Relevant equations
    Defn. of a combination.
    Defn. of mathematical induction.


    3. The attempt at a solution
    The formula is true for n=1
    2=∑j=01 n C r
    = 1 C 0 +1 C 1
    =1 + 1
    = 2
    Now assume that for some k∈ℕ and 0≤ j ≤ k we have
    2k = ∑kj=0 (k C j)
    Then
    (2)(2k)=2k+1= (2)kj=0 (k C j)

    LHS= ∑kj=0 (k C j) + ∑kj=0 (k C j)

    LHS= ∑kj=0 (k C j) + ∑k+1j=0 (k C j-1)

    LHS= ∑kj=0 (k C j) + ∑kj=0 (k C j-1) + (k+1 C k+1)

    LHS= ∑kj=0 [ (k C j) + (k Cj-1) ] +(k+1 C k+1)

    LHS= ∑kj=0 (k+1 C j) + (k+1 C k+1)

    LHS= ∑k+1j=0 (k+1 C j)

    Does everything check? In addition, how was this formula discovered? How do people even find these things? Thanks again for the help guys and gals.
     
    Last edited: Jun 10, 2015
  2. jcsd
  3. Jun 10, 2015 #2

    wabbit

    User Avatar
    Gold Member

    Seems OK, although it would be cleaner to adjust the ## j=0 ## starting point in some of your sums - these work if you define ## _k C _j= 0## for ## j=-1 ## but if you are using that you should say so.

    Otherwise I don't know how it was discovered but one way to see it directly is to note that ## _n C _k ## is the number of subsets of ## k ## elements of a set having ## n ## elements, and ## 2^n ## is the number of all subsets of a set having ## n ## elements.
     
    Last edited: Jun 10, 2015
  4. Jun 10, 2015 #3
    thanks for the reply. I forgot that nCr is the number of subsets of k elements of a set of n elements. I've seen other proofs for that.
     
  5. Jun 10, 2015 #4

    Ray Vickson

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    Science Advisor
    Homework Helper

    People discover these thing by first discovering the binomial theorem
    [tex] (1+x)^n = \sum_{k=0}^m {}_nC_k \, x^k [/tex]
    and then putting ##x = 1##.

    Newton discovered that result in 1665, before age 25.
     
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